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What is the heat released during the combustion of 5 grams of propane (C3H8) at constant volume, given that the enthalpy change of combustion of propane is -2220 kJ/mol?

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To calculate the heat released during the combustion of 5 grams of propane, we first need to determine the number of moles of propane in 5 grams. The molecular weight of propane  C3H8  is: 3 * 12.01  +  8 * 1.01  = 36.03 + 8.08 = 44.11 g/molNow, we can find the number of moles of propane in 5 grams:5 g / 44.11 g/mol = 0.1133 molesSince the enthalpy change of combustion of propane is -2220 kJ/mol, we can calculate the heat released during the combustion of 0.1133 moles of propane:Heat released =  0.1133 moles  *  -2220 kJ/mol  = -251.3 kJTherefore, the heat released during the combustion of 5 grams of propane at constant volume is approximately -251.3 kJ.

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